Taiwan IMO Selection Camp

(a)
Consider the number of inversions of an arrangement ($inv$); for example, the arrangement $23514$ has inversions $(2, 1)$, $(3, 1)$, $(5, 1)$, $(5, 4)$, so $inv = 4$.
After one operation, the part that changes the value of $inv$ only needs to consider:
$$\begin{aligned} & (a_j, a_{j+1}, *, a_k, a_{k+1}) \rightarrow (*, a_k, a_{k+1}, a_j, a_{j+1}) \\ & \rightarrow (a_k, a_{k+1}, *, a_j, a_{j+1}). \end{aligned}$$
The intermediate step is intentionally inserted. For any $a_i$ among $(*, a_k, a_{k+1})$, moving $a_j$ to behind it will increase or decrease $inv$ by $1$; moving $a_{j+1}$ to behind it is similar; so moving $a_j, a_{j+1}$ with respect to $a_i$ changes $inv$ by $\pm 2$ or leaves it unchanged; therefore, for all the numbers in $(*, a_k, a_{k+1})$, moving $a_j, a_{j+1}$ changes $inv$ by an even amount. This completes the first step, and the second step is similar: for all the numbers in $*$, moving $a_k, a_{k+1}$ to the front also changes $inv$ by an even amount. The conclusion is:
After each operation, the parity of $inv$ does not change.
Clearly, $1, 2, \dots, n$ has $inv = 0$. And $n, n-1, \dots, 1$ has $inv = (n-1) + (n-2) + \dots + 1 = n(n-1)/2$, which is even if and only if $n \equiv 0, 1 \pmod 4$.

Since $n = 2001 \equiv 1 \pmod 4$, both the increasing and decreasing arrangements have even $inv$, so it is clear that the arrangement $2, 1, 3, 4, \dots, n$ with $inv = 1$ can never, no matter how many operations are performed, become either the increasing or decreasing arrangement.

(b)
$n = 2011 \equiv 3 \pmod 4$, so the $inv$ of $n, n-1, \dots, 1$ is odd, so it is possible.

Theorem If $n \ge 5$ and $n \equiv 2,3 \pmod 4$, then any arrangement with even (odd) $inv$ can be transformed into the increasing (decreasing) arrangement.

Proof
We first prove the case "when $inv$ is even". Using positions $1,2,3,4$ as a buffer, place $n, n-1, \dots, 5$ into their proper positions in order; for example for $5$ (the following is not the most efficient, but it lets $5$ appear respectively in positions $1,3,4$; the case where $5$ appears in position $2$ is simple enough to need no explanation):
$$(a, b, c, 5, d, 6, 7, \dots) \rightarrow (5, d, c, a, b, 6, 7, \dots) \rightarrow (c, a, 5, d, b, 6, 7, \dots) \rightarrow (c, d, b, a, 5, 6, 7, \dots).$$
Keeping $(6,7,\dots,n)$ unchanged, we now need to handle all arrangements $(A,B,C,D,5)$ with even $inv$, where $\{A,B,C,D\} = \{1,2,3,4\}$. Now the arrangements of $1,2,3,4$ number $24$ in total, and those with even $inv$ account for half, namely $12$, listed as follows:
(i) $(1,2,3,4), (1,3,4,2), (1,4,2,3)$ fix $1$, then $2,3,4$ cycle;
(ii) $(2,1,4,3), (2,4,3,1), (2,3,1,4)$ fix $2$ in the first position, then cycle;
(iii) $(3,1,2,4), (3,2,4,1), (3,4,1,2)$ fix $3$ in the first position, then cycle;
(iv) $(4,1,3,2), (4,3,2,1), (4,2,1,3)$ fix $4$ in the first position, then cycle.

Define two combined methods of operation:
$$(\alpha) : (A, B, C, D, 5) \rightarrow (D, 5, C, A, B) \rightarrow (D, A, B, 5, C) \\ \rightarrow (B, 5, D, A, C) \rightarrow (A, C, D, B, 5)$$
This amounts to fixing $A$, then cycling $B, C, D$.
$$(\beta) : (A, B, C, D, 5) \rightarrow (A, D, 5, B, C) \rightarrow (B, C, 5, A, D) \\ \rightarrow (B, A, D, C, 5)$$
This swaps the original $A, B$, and also swaps $C, D$.
Operation $(\alpha)$ can link the three members within each of groups (i)–(iv) to each other. Via $\beta$ we have
$$(2, 1, 4, 3) \rightarrow (1, 2, 3, 4);$$
$$(3, 1, 2, 4) \rightarrow (1, 3, 4, 2);$$
$$(4, 1, 3, 2) \rightarrow (1, 4, 2, 3)$$
The proof of "when $inv$ is even" is complete!

For the proof of "when $inv$ is odd", we can change every number $a_i$ to $a_i' = n + 1 - a_i$, at which point the new $inv$ becomes $n(n-1)/2$ minus the old $inv$; because $n \equiv 1,3 \pmod 4$, the new $inv$ is even; following the previous proof we can obtain the arrangement $1,2,...,n$, and then restoring all the numbers ($k \rightarrow n+1-k$) turns it into $n,n-1,...,1$. Proof complete!